SOLUTION 4: We are given the function $f(x) = \displaystyle{ x \over 1+x }$ and the interval $[1, 3]$. This function is continuous on the closed interval $[1, 3]$ since it is the quotient of continuous functions $y=x$ (polynomial) and $y=1+x$ (polynomial). The derivative of $f$ is $$f'(x) = \displaystyle{ (1+x)(1) - x(1) \over (1+x)^2 } = { 1 \over (1+x)^2 }$$ We can now see that $f$ is differentiable on the open interval $(1, 3)$. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of $c$ in the open interval $(1, 3)$. Then $$f'(c)= \displaystyle{ f(3)-f(1) \over 3-1 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ { (3) \over 1+(3) } - { (1) \over 1+(1) } \over 3-1 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ { 3 \over 4 } - { 1 \over 2 } \over { 2 \over 1 } } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ \Big( { 3 \over 4 } - { 2 \over 4 } \Big) \cdot { 1 \over 2 } } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over (1+c)^2 } = \displaystyle{ 1 \over 8 } \ \ \ \ \longrightarrow$$ $$(1+c)^2 = 8 \ \ \ \ \longrightarrow$$ $$1+c = \pm \sqrt{8} = \pm 2 \ \sqrt{2} \ \ \ \ \longrightarrow$$ $$c = \pm 2 \ \sqrt{2} -1 \ \ \ \ \longrightarrow$$ $$c = 2 \ \sqrt{2} -1 \approx 1.828 \ \ \ or \ \ \ -2 \ \sqrt{2} -1 \approx -3.828 \ \ \ (-3.828 \ is \ NOT \ in \ the \ interval \ (1, 3).) \ \ \ \ \longrightarrow$$ $$c = 2 \ \sqrt{2} -1 \approx 1.828 \ \ \ \ \longrightarrow$$